Question

The figure shows a kite ABCD where AB =AD,BC = CD and the diagonals AC and BD intersect at E
DAE=25
DCE=44
find ABD and CBD

Answer 1

Answer:

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 (90o 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:

The diagonals intersect at right angles

Diagonal AC bisects diagonal BD

Diagonal AC bisects angles BAD and BCD

Given property 3 above, if angle CAD = 40o then angle BAC = 40o

Look at triangle BCE. Angle CBE is given as 60o and angle BEC = 90o (property 1 above).

Angle BCA (same as BCE) = 180o - (90o+60o) = 30o

By property 3, if angle BCA = 30o, then angle BCD = 60o