(a) If a circle is inscribed in a right angled triangle ABC with the right angle at B, show that the

Q: Wait, why does $r = s - b$ only work for right triangles?

Actually, the formula $r = (s - b) \tan(\frac{B}{2})$ is a general formula for any triangle. In a right-angled triangle, $\angle B = 90^\circ$, so $\tan(\frac{90^\circ}{2}) = \tan(45^\circ) = 1$. That is why it simplifies so beautifully to $r = s - b$ specifically for right triangles. It is not that the formula is restricted; it is just that the geometry of the right angle makes the trigonometry trivial!

Q: Why is $\angle ABE = 90^\circ$ in the second part?

That is a classic circle theorem! Any angle subtended by a diameter at the circumference of a circle is always a right angle. Since $AE$ is the diameter, the triangle $ABE$ is inscribed in a semi-circle, forcing $\angle ABE$ to be $90^\circ$. It is a powerful property that often unlocks geometry problems involving circumcircles.

Q: Can we use the area method for the first part instead?

Absolutely! You can use the area formula $Area = rs$, where $s$ is the semi-perimeter. For a right triangle, the area is also $\frac{1}{2} \times a \times c$. By equating $\frac{1}{2}ac = r(\frac{a+b+c}{2})$, you can solve for $r$. Both methods are elegant, but the $r = s - b$ approach is often faster for right triangles. Keep exploring both!

Question

(a) If a circle is inscribed in a right angled triangle $ABC$ with the right angle at $B$, show that the diameter of the circle is equal to $AB + BC - AC$. (b) If a triangle is inscribed in a circle, then the product of any two sides of the triangle is equal to the product of the diameter and the perpendicular distance of the third side from the opposite vertex. Prove the above statement.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a right-angled triangle $ABC$, with the right angle at $B$. We label the sides opposite to the vertices as $a, b,$ and $c$.

Specifically, we define $BC = a$, $AC = b$ (the hypotenuse), and $AB = c$. We place a circle inside this triangle, touching all three sides, known as the incircle with radius $r$.

Our mission is to prove that the diameter $d = 2r$ is equal to $AB + BC - AC$, or $a + c - b$.

### The Inradius Derivation

We utilize the general inradius formula: 
$$r = (s - b) 	an\left(\frac{B}{2}ight)$$ 
where $s$ is the semi-perimeter, defined as $s = \frac{a + b + c}{2}$.

Because our triangle is right-angled at $B$, we know $\angle B = 90^\circ$. Substituting this into our formula, we get:
$$r = (s - b) 	an(45^\circ)$$

Since $	an(45^\circ) = 1$, the expression simplifies elegantly to $r = s - b$. Substituting the definition of $s$ into this equation yields:
$$r = \frac{a + b + c}{2} - b$$

By finding a common denominator, we obtain:
$$r = \frac{a + b + c - 2b}{2} = \frac{a + c - b}{2}$$

Multiplying by $2$, we find $2r = a + c - b$. Since $2r$ is the diameter, we have proven that the diameter is indeed $AB + BC - AC$.

### The Triangle in the Circle

Now, let us shift our perspective to a triangle $ABC$ inscribed in a circle. We drop an altitude $AD$ from vertex $A$ to the base $BC$ and draw a diameter $AE$ passing through the center.

We aim to prove the relationship $AB 	imes AC = AE 	imes AD$. This product of sides suggests the existence of similar triangles.

To visualize this, we construct a line connecting $B$ to $E$. Now, consider $	riangle ABE$ and $	riangle ADC$.

In $	riangle ABE$, $\angle ABE = 90^\circ$ because it is an angle inscribed in a semi-circle. In $	riangle ADC$, $\angle ADC = 90^\circ$ by construction. We have identified our first pair of equal angles.

### Establishing Similarity

Next, consider the arc $AB$. It subtends $\angle AEB$ and $\angle ACB$ (which is $\angle ACD$) at the circumference.

By the theorem that angles in the same segment are equal, we know $\angle AEB = \angle ACD$. With two pairs of equal angles, we have established that $	riangle ABE \sim 	riangle ADC$ by the **Angle-Angle (AA) similarity criterion**.

### Final Calculation

Because the triangles are similar, the ratios of their corresponding sides must be equal:
$$\frac{AB}{AD} = \frac{AE}{AC}$$

Cross-multiplying these terms gives us the final result:
**$$AB 	imes AC = AE 	imes AD$$**

This is the beauty of geometry—a complex relationship between lengths is reduced to a simple, elegant product. The underlying mathematical laws remain perfectly consistent, whether we are analyzing the circle inside or the circle outside the triangle.

Visualized Solution

Visualizing the Inscribed Circle

The Inradius Formula

Substituting the Angle

Simplifying the Expression

Final Diameter Relation

Part (b): Triangle in a Circle

Identifying Right Angles

Angles in the Same Segment

The Final Ratio

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Inradius Derivation

The Triangle in the Circle

Establishing Similarity

Final Calculation

Similar Questions

In a triangle $A B C$ , let $∠ C = \frac{π}{2}$ . If $r$ is the inradius and $R$ is the circumradius of the triangle $A B C$ , then $2 (r + R)$ equals

$A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that $cos A cos C = \frac{2 ( c ^{2} - a ^{2} )}{3 a c}$ .

Let $a, b$ and $c$ be the length of sides of a triangle $A B C$ such that $\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9}$ . If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $A B C$ , respectively, then the value of $\frac{R}{r}$ is equal to

In a triangle $A B C$ , let $∠ C = π /2$ . If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2 (r + R)$ is equal to

In a triangle the sum of two sides is $x$ and the product of the same sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

A circle is inscribed in an equilateral triangle of side $a$ . The area of any square inscribed in this circle is ..................

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .