Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

Suppose $A B C D$ is a quadrilateral such that a semicircle with its centre at the midpoint of $A B$ and bounding diameter lying on $A B$ touches the other three sides $B C, C D$ and $D A$ . Show that $A B^{2} = 4 B C \cdot A D$ .

Accepted Answer

Let $K, L, M$ be the points of contact of the semicircle with the sides $B C, C D, D A$ respectively.

Join $O K, O L, O M, O C$ and $O D$ , where $O$ is the centre of the circle as well as the midpoint of $A B$ .

Sketching the diagram :

In triangles $A O M$ and $B O K$ , we have

$A O = B O$ (given)

$∠ A M O = ∠ B K O = 90 °$ (line joining centre to the point of contact of tangent is perpendicular to each other)

$O M = O K$ (radii are equal)

So, $△ A O M ≅ △ B O K$ by RHS congruency.

Thus, corresponding parts of congruent triangles are equal.

i.e., $∠ A O M = ∠ B O K = α$ (let say) and $∠ O A M = ∠ O B K . . . . . . (i)$

Now, in triangles $O M D & O L D$ , we have

$O M = O L$ (radii are equal)

$∠ O M D = ∠ O L D = 90 °$ (line joining centre to the point of contact of tangent is perpendicular to each other)

$O D = O D$ (common)

So, $△ O M D ≅ △ O L D$ by RHS congruency.

Thus, corresponding parts of congruent triangles are equal.

i.e., $∠ D O M = ∠ D O L = β . . . . . . . (ii)$ (let say)

Similarly, $△ O K C ≅ △ O L C$ by RHS congruency.

So, $∠ C O L = ∠ C O K = γ$ (let say)

Adding up the angles formed at $O$ , we obtain

$2 α + 2 β + 2 γ = 180 °$

$\Rightarrow α + β + γ = 90 ° . . . . . . (iii)$

Now, in triangles $A O D$ and $B C O$ , we have

$∠ O A D = 90 ° - ∠ A O M = 90 ° - α$ and $∠ C B O = 90 ° - ∠ K O B = 90 ° - α$

Also, $∠ A O D = α + β = 90 ° - γ$ and $∠ B C O = 90 ° - ∠ C O K = 90 ° - γ$

So, $△ A O D ~ △ B C O$ by AA-similarity.

Thus, $\frac{A O}{A D} = \frac{B C}{B O}$ .

$\Rightarrow A D \cdot B C = A O \cdot B O = \frac{1}{2} A B \cdot \frac{1}{2} A B = \frac{1}{4} A B^{2}$

$\Rightarrow A B^{2} = 4 A D \cdot B C$

Hence, proved.

Suppose $A B C D$ is a quadrilateral such that a semicircle with its centre at the midpoint of $A B$ and bounding diameter lying on $A B$ touches the other three sides $B C, C D$ and $D A$ . Show that $A B^{2} = 4 B C \cdot A D$ .

Important Questions on Geometry

Learn and Prepare for any exam you want !

Suppose ABCD is a quadrilateral such that a semicircle with its centre at the midpoint of AB and bounding diameter lying on AB touches the other three sides BC,CD and DA. Show that AB2=4BC·AD.

Important Questions on Geometry

Learn and Prepare for any exam you want !

Suppose $A B C D$ is a quadrilateral such that a semicircle with its centre at the midpoint of $A B$ and bounding diameter lying on $A B$ touches the other three sides $B C, C D$ and $D A$ . Show that $A B^{2} = 4 B C \cdot A D$ .