Question
In the given figure, $AD = AE$ and $\angle BAD =\angle CAE$. Prove that: $AB = AC$.
✓
Answer
Hint :
[Prove that $\triangle A B D \cong \triangle A C E$.$\angle A D E=\angle A E D \Rightarrow 180^{\circ}-\angle A D E=180^{\circ}-\angle A E D$
$\Rightarrow \angle A D B=\angle A E C .]$
[Prove that $\triangle A B D \cong \triangle A C E$.$\angle A D E=\angle A E D \Rightarrow 180^{\circ}-\angle A D E=180^{\circ}-\angle A E D$
$\Rightarrow \angle A D B=\angle A E C .]$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate :$\left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$
→Solve for $\mathrm{x}$ and $\mathrm{y}$; if $\mathrm{x}>0$ and $\mathrm{y}>0 ; \log \mathrm{xy}=\log \frac{x}{y}+2 \log 2=2$
→The radii of two circles are $48 \ cm$ and $13 \ cm$. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.
→Simplify the following:$(81)^{\frac{3}{4}}-\left(\frac{1}{32}\right)^{-\frac{2}{5}}+8^{\frac{1}{3}} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 2^0$
→The diagonals of a rectangle intersect each other at right angles. Prove that the rectangle is a square.
Divide $Rs. 28,730$ between $A$ and $B$ so that when their shares are lent out at $10\%$ compound interest compounded per year, the amount that $A$ receives in $3$ years is the same as what $B$ receives in $5$ years.
→Factorise the following:$9(a - b)^2 - (a + b)^2$
→An isosceles $\triangle ABC$ has $AC = BC. CD$ bisects $AB$ at $D$ and $\angle CAB = 55^o$.Find:$(i)\angle DCB(ii)\angle CBD.$
→Construct a right$-$angled triangle in which Side $DE = 6 \ cm$ and $\angle E = 30^\circ , \angle D = 90^\circ $
→Use the information given in the following figure to evaluate:$\frac{10}{\sin x}+\frac{6}{\sin y}-6 \cot y$
→