Question

In parallelogram ABCD, P is a point on BC. In ∆ DCP and ∆ BLP, DP:PL is equal to?

Answer 1

Answer:

Solution:

Let ABCD be a parallelogram and L extended to prove that \frac{D P}{P L}=\frac{D C}{B L}

PL

DP

=

BL

DC

Let us first show that in triangle PCD and BPL, the angle of DPC and BPL are same as they are vertically opposite to each other.

The angle C = B, as they are alternate angle, therefore it shows that triangle PCD and BPL are similar in nature, therefore sides DP = PL and DC = BL.

\bold{\frac{D P}{P L}=\frac{D C}{B L}}

PL

DP

=

BL

DC

Hence Proved

Now it can be said that as AB = DC, due to parallel sides of a parallelogram, it is known that

\frac{D P+P L}{D P}=\frac{A B+B L}{D C}

DP

DP+PL

=

DC

AB+BL

Hence, \bold{\frac{D L}{D P}=\frac{A L}{D C}}

DP

DL

=

DC

AL

Proved

Step-by-step explanation:

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