Question

If ABCD is a rhombus in which the altitude from D to side AB bisects AB. Then $\angle A$ and $\angle B$ respectively, are...
$\left( A \right){60^o},{120^o}$
$\left( B \right){120^o},{60^o}$
$\left( C \right){80^o},{100^o}$
$\left( D \right){100^o},{80^o}$

Answer 1

Hint: In this particular type of question use the concept that if two triangles are similar then the ratio of the respective side lengths is always equal and in an equilateral triangle all angles are equal to 60 degrees so use these concepts to reach the solution of the question.

Complete step-by-step answer:

ABCD rhombus is shown above.
Now it is given that the altitude is drawn from point D on line AB and it bisects the line AB.
Let it intersect the line AB at point P.
Therefore, PA = PB.
Now as we all know that the altitude is nothing but the perpendicular.
Therefore, $ \Rightarrow \angle DPA = \angle DPB = {90^o}$
So in triangle ADP and BDP we have,
PA = PB
$\angle DPA = \angle DPB = {90^o}$
DP = DP (common)
So by SAS congruence the triangles ADP and BDP are similar to each other.
Therefore, AD = BD.............. (1)
Now as we know that in rhombus all sides are equal.
Therefore, AD = AP................ (2)
Now from equation (1) and (2) we have,
AD = AP = BD
So the triangle ADB is an equilateral triangle.
As we all know that in an equilateral triangle all angles are equal and = 60 degrees.
$ \Rightarrow \angle DAB = \angle DBA = \angle ADB = {60^o}$
Therefore, $\angle A = {60^o}$
Now, $\angle DBA = \angle DBC = {60^o}$
So, $\angle B = \angle DBA + \angle DBC = {60^o} + {60^o} = {120^o}$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that in a rhombus all sides are equal and the diagonal of the rhombus always bisect the respective angle of the diagonal so first prove that the triangles ADP and BDP are similar to each other then apply rhombus condition so that the big triangle is an equilateral triangle.