In quadrilateral ABCD, BM and DN are drawn perpendiculars to AC such that BM = DN. If BR = 8 cm, then find BD.

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Solution

Given $A B C D$ is a quadrilateral in which diagonals $A C$ and $B D$ intersect each other at $P$
Also, $D N ⊥ A C$ and $B M ⊥ A C$
$D N = B M$
In $△ D N P$ and $△ B M P$ we have
$∠ D P N = ∠ B P M$ since they are vertically opposite angles.
$∠ D N P = ∠ B M P$ since each are equal to $90^{\circ}$
$D N = B M$ (given)
$\Rightarrow △ D N P ≅ △ B M P$ by AAS postulate.
$\Rightarrow D P = B P = 8$ cms by CPCT
We have $B D = B P + D P = 8 + 8 = 16$ cms from the diagram.

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In Fig. BM and DN are both perpendiculars to the segments AC and BM = DN. Prove that AC bisects BD [3 MARKS]

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