Question

In the given figure, $∠ A = ∠ D = 90 °$ , $∠ C = 48 °$ . $B E$ is bisector of $∠ B$ , $A D$ and $B E$ intersect at $M$ . Calculate prove that $A E = A M$ .

Accepted Answer

In a triangle $A B C$ ,

$∠ C A B + ∠ A B C + ∠ A C B = 180 °$ $($ Angle sum property $)$

$\Rightarrow 90 ° + 48 ° + ∠ A B C = 180 °$

$\Rightarrow ∠ A B C = 180 ° - 138 ° = 42 °$

$∴ ∠ E B A = \frac{∠ A B C}{2} = \frac{42 °}{2} = 21 °$ $($ $B E$ is the bisector of $∠ A B C)$

Now, in $∆ M D B$

$∠ M D B + ∠ D B M + ∠ B M D = 180 °$ $($ Angle sum property $)$

$\Rightarrow 90 ° + 21 ° + ∠ B M D = 180 °$

$\Rightarrow 111 ° + ∠ B M D = 180 °$

$\Rightarrow ∠ B M D = 180 ° - 111 °$

$\Rightarrow ∠ B M D = 69 °$

Since, $∠ A M E = ∠ B M D = 69 °$ $($ Vertically opposite angles $)$

Now, in $∆ E A B$ ,

$∠ E A B + ∠ A B E + ∠ A E B = 180 °$ $($ Angle sum property $)$

$\Rightarrow 90 ° + 21 ° + ∠ A E B = 180 °$

$\Rightarrow ∠ A E B = 180 ° - 111 ° = 69 °$

So, $∠ A E M = ∠ A M E$

We know that sides opposite to equal angles are equal.

Hence, $A E = A M$

K C Sisodia Solutions for Chapter: Geometry, Exercise 1: TEST YOUR SKILLS

K C Sisodia Mathematics Solutions for Exercise - K C Sisodia Solutions for Chapter: Geometry, Exercise 1: TEST YOUR SKILLS

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