Question

In quadrilateral ABCD, BN and DM are
drawn perpendicular to AC. Such that
BN =Dm. Prove that o is mid-point of BD.​

Answer 1

Step-by-step explanation:

In DOM AND BON

<OMD =<ONB (90°)

DM = BN (given)

<DOM = <BON (V O <s)

DOM is congruent to BON ( BY AAS)

DO =OB ( by CPCT)

therefore o is mid point of BD

hence proved

Answer 2

DO = OB , this means O is mid point of BD

Hence proved

Step-by-step explanation:

In the given figure, ABCD is a quadrilateral.

BN and DM are  drawn perpendicular to AC such that BN = DM

In ΔDMO and ΔBNO

 ∠DMO = ∠BNO   ( each 90°)

 ∠DOM = ∠BON   (vertically opposite angle)

∵ ΔDMO ≈ ΔBNO  by AA similarity

If two triangle are similar then their sides are in proportion.

$\dfrac{MO}{NO}=\dfrac{DO}{BO}=\dfrac{DM}{BN}$

$\dfrac{DO}{BO}=\dfrac{BN}{BN}$        $\because BN = DM$

$\dfrac{DO}{BO}=1$

$DO=BO$

This means O is mid point of BD.

Hence prove

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