In the figure, $ABC$ is a right triangle right angled at $A, BCED, ACFG$ and $ABMN$ are squares on the sides $BC, CA$ and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that $\triangle MBC \cong \triangle ABD$.
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Given:

$ABC$ is a right triangle right angled at $A, BCED, ACFG$ and $ABMN$ are squares on the sides $BC, CA$ and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$.

To do:

We have to show that $\triangle MBC \cong \triangle ABD$.

Solution:

In $\triangle MBC$ and $\triangle ABD$,

$MB=AB$            (Sides of square)

$BC = BD$

$\angle MBC = \angle ABD = 90^o + \angle ABC)$

Therefore, by SAS axiom,

$\triangle MBC \cong \triangle ABD$

This implies,

$ar(\triangle MBC) = ar(\triangle ABD)$........…(i)

Hence proved.

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Updated on: 10-Oct-2022

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