Question
In the given figure, $AD = AE$ and $\angle BAD =\angle CAE$. Prove that: $AB = AC$.
Image

Answer

self

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

ABCD is a rectangle in which diagonal BD bisects$\angle B$. Can we definitely say that ABCD is a square?
Insert a rational number between:$\frac{3}{4}$ and $\frac{5}{7}$
By converting to exponential form, find the value of the following :
$\log _2 64$
If $a=\frac{1}{(a-5)}$, where $a \neq 5$ and $a \neq 0$, find the values of :
(i) $\left(a-\frac{1}{a}\right)$ (ii) $\left(a+\frac{1}{a}\right)$ (iii) $\left(a^2-\frac{1}{a^2}\right)$ (iv) $\left(a^2+\frac{1}{a^2}\right)$
[Hint. $a=\frac{1}{a-5} \Rightarrow a-5=\frac{1}{a} \Rightarrow\left(a-\frac{1}{a}\right)=5$.]
Factorise the following by splitting the middle term:$7x^2 + 40x - 12$
In a parallelogram $\text{ABCD}, E$ and $F$ are the midpoints of the sides $AB$ and $CD$ respectively. The line segments $AF$ and $BF$ meet the line segments $DE$ and $CE$ at points $G$ and $H$ respectively Prove that$: \triangle GEA \cong \triangle GFD$
$\text{ABCDE}$ is the end view of a factory shed which is $50\ m$ long. The roofing of the shed consists of asbestos sheets as shown in the figure. The two ends of the shed are completely closed by brick walls.
Image
Calculate the total volume content of the shed.
Without using tables, find the value of the following: $\frac{\sin 30^{\circ}}{\sin 45^{\circ}}+\frac{\tan 45^{\circ}}{\sec 60^{\circ}}-\frac{\sin 60^{\circ}}{\cot 45^{\circ}}-\frac{\cos 30^{\circ}}{\sin 90^{\circ}}$
The mean of $4$ observations is $20.$ If one observation is excluded, the mean of the remaining observations becomes $15.$ Find the excluded observation.