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Question

Show that the bisector of angles of a parallelogram from a rectangle.


Solution

Let a parallelogram $$ABCD$$

we know that

$$\angle PAS=\angle RAB=\dfrac{1}{2}<  PAB$$..............(1)
Similarly

$$\angle CBQ=\angle QBA=\dfrac{1}{2}<  CBA$$..............(2)
So

$$AD||BC$$ then

$$\angle OAB+\angle CAB=180^0$$

$$\dfrac{1}{2}\angle DAB+\dfrac{1}{2}\angle CAB=90^0$$

$$\angle RAB+\angle RBA=180^0$$

Now In $$\Delta le$$ $$ RAB$$

$$\angle RAB+\angle RBA_\angle ARB=180^0$$

$$90^0+\angle ARB=180^0$$

$$\boxed{\angle ARB=90^0}$$

Similarly

$$\boxed{\angle DPC=90^0}$$

So opposite angles are $$90^0$$, so

It is a rectangle.

The first line of solution should be

$$ \angle DAS = \angle RAB = \dfrac{1}{2} = \angle DAB $$
1950579_1230383_ans_17902702f684433f91d8615aedfe6a10.png

Maths

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