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In figure below, $ \mathrm{ABC} $ is a right triangle right angled at $ \mathrm{A} . \mathrm{BCED}, \mathrm{ACFG} $ and $ \mathrm{ABMN} $ are squares on the sides $ \mathrm{BC}, \mathrm{CA} $ and $ \mathrm{AB} $ respectively. Line segment $ \mathrm{AX} \perp \mathrm{DE} $ meets $ \mathrm{BC} $ at Y. Show that:
(i) $ \triangle \mathrm{MBC} \cong \triangle \mathrm{ABD} $
(ii) $ \operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC}) $
(iii) $ \operatorname{ar}(\mathrm{BYXD})=\operatorname{ar}(\mathrm{ABMN}) $
(iv) $ \triangle \mathrm{FCB} \cong \triangle \mathrm{ACE} $
(v) $ \operatorname{ar}(\mathrm{CYXE})=2 \operatorname{ar}(\mathrm{FCB}) $
(vi) $ \operatorname{ar}(\mathrm{CYXE})=\operatorname{ar}(\mathrm{ACFG}) $
(vii) ar $ (\mathrm{BCED})=\operatorname{ar}(\mathrm{ABMN})+\operatorname{ar}(\mathrm{ACFG}) $"
Given:
\( \mathrm{ABC} \) is a right triangle right angled at \( \mathrm{A} . \mathrm{BCED}, \mathrm{ACFG} \) and \( \mathrm{ABMN} \) are squares on the sides \( \mathrm{BC}, \mathrm{CA} \) and \( \mathrm{AB} \) respectively. Line segment \( \mathrm{AX} \perp \mathrm{DE} \) meets \( \mathrm{BC} \) at Y.
To do:
We have to show that
(i) \( \triangle \mathrm{MBC} \cong \triangle \mathrm{ABD} \)
(ii) \( \operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC}) \)
(iii) \( \operatorname{ar}(\mathrm{BYXD})=\operatorname{ar}(\mathrm{ABMN}) \)
(iv) \( \triangle \mathrm{FCB} \cong \triangle \mathrm{ACE} \)
(v) \( \operatorname{ar}(\mathrm{CYXE})=2 \operatorname{ar}(\mathrm{FCB}) \)
(vi) \( \operatorname{ar}(\mathrm{CYXE})=\operatorname{ar}(\mathrm{ACFG}) \)
(vii) ar \( (\mathrm{BCED})=\operatorname{ar}(\mathrm{ABMN})+\operatorname{ar}(\mathrm{ACFG}) \)
Solution:
In $\triangle ABD$ and $\triangle MBC$,
$BC = BD$ (Sides of a square are equal)
$MB = AB$
$\angle MBC = 90^o + \angle ABC$
$= \angle DBC + \angle ABC$
$= \angle ABD$
Therefore, by SAS congruency,
$\triangle MBC = \triangle ABD$
(ii) $ar(\triangle MBC) = ar (\triangle ABD)$.......…(i)
$ar(\triangle ABD) = \frac{1}{2}ar (BYXD) …(ii) (Since $\triangle ABD$ and rectangle $BYXD$ lie on the same base and between the same parallels)
From (i) and (ii), we get,
$ar (\triangle MBC) = \frac{1}{2}ar (BYXD)$......(iii)
$ar (BYXD) = 2 ar (\triangle MBC)$
(iii) $ar (\triangle MBC) = \frac{1}{2} ar (ABMN)$…..(iv) (SInce $\triangle MBC$ and square $ABMN$ lie on the same base $MB$ and between the same parallels $MB$ and $NC$)
From (iii) and (iv), we get,
$ar (BYXD) = ar (ABMN)$
(iv) In $\triangle ACE$ and $\triangle FCS$,
$AC = FC$
$CE = BC$ (Sides of the square)
$\angle FCB = 90^o + \angle ACB$
$= \angle BCE + \angle ACB$
$= \angle ACE$
Therefore, by SAS congruency,
$\triangle FCB = \triangle ACE$
(v) $ar(\triangle ACE) = ar(AFCB)$....…(vi)
$ar(\triangle ACE) = \frac{1}{2} ar(CVXE)$ (Since both lie on the same base $CE$ and between the same parallels $CE$ and $AX$)
From (vi) and (vii), we get,
$ar (\triangle ACE) = \frac{1}{2} ar (CYXE)$
$= ar (\triangle FCB)$.....…(vii)
$ar (CYXE) = \frac{1}{2} ar (\triangle FCB)$
(vi) $ar(AFCB) = \frac{1}{2} ar (ACFG)$......…(ix) (Since both lie on the same base $CF$ and between the same parallels $CF$ and $BG$)
From (viii) and (ix), we get,
$\frac{1}{2} ar (ALFG) = \frac{1}{2} ar (CYXE)$
$ar (ACFG) = ar (CYXE)$
(vii) $ar (BCED) = ar (BYXD) + ar (CYXE)$
$= ar (ABMN) + ar (ACFG)$ [From(iii) and (vi)]
Hence proved.
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