Question
Observe the Fig. given below and then find $\angle$P.
✓
Answer
Given, $\triangle$ABC $\sim\triangle$FEG ….(1)
(i) Corresponding angles of similar triangles
$\Rightarrow$ $\angle$BAC = $\angle$EFG ….(2)
And $\angle$ABC = $\angle$FEG …(3)
$\Rightarrow$ $\angle$ACB = $\angle$FGE
$\Rightarrow \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE$
$\Rightarrow$ $\angle$ACD = $\angle$FGH and $\angle$BCD = $\angle$EGH ……(4)
Consider $\triangle$ACD and $\triangle$FGH
$\Rightarrow$ From (2) we have
$\Rightarrow$$\angle$DAC = $\angle$HFG
From (4) we have
$\Rightarrow$ $\angle$ACD = $\angle$EGH
Also, $\angle$ADC = $\angle$FGH
If the $\angle A=\angle F$, then by angle sum property of triangle $3^{rd}$ angle will also be equal.
By AAA similarity, in two triangles, if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar.
$\therefore\triangle$ADC $\sim\triangle$FHG
(ii) By Converse proportionality theorem
$\Rightarrow \frac{C D}{G H}=\frac{A C}{F G}$
(iii) Consider $\triangle$DCB and $\triangle$HGE
From eq(3) we have
$\Rightarrow$ $\angle$DBC = $\angle$HEG
From (4) we have
$\Rightarrow$ $\angle$BCD = $\angle$FGH
Also, $\angle$BDC = $\angle$EHG
$\therefore\triangle$DCB $\sim\triangle$HGE
Hence proved.
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