Question

5. In the adjoining figure, BM LAC and
DN 1 AC. If BM = DN, prove that AC
bisects BD.​

Answer 1

Solution:

Note: Diagram of this question attached in attachment file.

Given: A quadrilateral ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.

To prove: AC bisects BD or DO = BO.

Proof:

Let AC and BD intersect at O.

Now, in ∆OND and ∆OMB,

we have:

∠OND = ∠OMB                 (90° each)

∠DON = ∠ BOM                (Vertically opposite angles)

Also, DN = BM                   (Given)'

i.e.,

∆OND ≅ ∆OMB             (AAS congurence rule)

∴ OD = OB                      (CPCT)

​Hence, AC bisects BD.

Answer 2

ANSWER:-

Given:

In figure, BM ⊥AC & DN⊥AC.If BM= DN.

To prove:

Prove that AC bisect BD.

Proof:

BR = RD

In ∆BMR & ∆DNR,

∠BMR = ∠DNR [each 90°]

∠BRM = ∠DRN [Vertically opposite ∠]

BM = DN [given]

So,

∆BMR ≅∆DNR [AAS congruence rule]

BR = RD [c.p.c.t]

Hence,

AC bisect BD.

Hope it helps ☺️