Question
In the given figure : $\angle BAC =\angle CDB$ and $\angle BCA =\angle CBD$. Prove that $A B=C D$.
✓
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.
Explore more
Similar questions
Express each of the following in terms of trigonometric ratios of angles between $0^\circ $and $45^\circ : \sin65^\circ + \cot59^\circ $
→Write down the values of:
$(3+2 \sqrt{5})^2$
→$(3+2 \sqrt{5})^2$
If $\log 27=1.4313$, find the value of: $\log 30$
→Find the mean of all factors of $10.$
→The simple interest on a sum of money is the product of the sum of money, the number of years and the rate percentage. Write the formula to find the simple interest on Rs $A$ for $T$ years at $R%$ per annum.
→Find each exterior angle of a regular polygon of $: 9$ sides
→Find the product using suitable identities : (ii) $\left(y-\frac{6}{y}\right)\left(y^2+6+\frac{36}{y^2}\right)$
→Define the following term:
Class-interval
Class-interval
Construct a regular hexagon of side 3.5 cm , using ruler and compasses only.
Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD . Prove that :
(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.
→(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.