Question

ABCD is a quadrilateral. The diagonals intersect at R. BM and DN are perpendiculars drawn from B and D on the diagonal AC. Prove that AC bisects BD.​

Answer 1

Answer:

BMR ≅ ΔDRN

BD =

Step-by-step explanation:

in ΔBMR & ΔDRN

∠BMR = ∠DNR = 90°

∠MRB = ∠NRD ( opposite angles)

if two angles of a triangle are equal then third angle will also be equal

as sum of three angles = 180°

∠MBR = ∠NDR

so

ΔBMR ≅ ΔDRN

BM/DN = BR/DR = MR/RN

BM = DN => BM/DN = 1

=> 1 = BR/DR

=> DR = BR

=>

Answer 2

Step-by-step explanation:

Answer

GivenABCD is a quadrilateral in which diagonals AC and BD intersect each other at P

Also,DN⊥AC and BM⊥AC

DN=BM

In △DNP and △BMP we have

∠DPN=∠BPM since they are vertically opposite angles.

∠DNP=∠BMP since each are equal to 90

∘

DN=BM(given)

⇒△DNP≅△BMP by AAS postulate.

⇒DP=BP=8cms by CPCT

We have BD=BP+DP=8+8=16cms from the diagram.