Question

In the figure, given below, AM bisects angle
A and DM bisects angle D of parallelogram
ABCD. Prove that : angle AMD = 90°.​

Answer 1

Answer:

ABCD is a parallelogram.

We know, the sum of all angles of a parallelogram is 360°,

∴ Each angle of the parallelogram = 360° / 4 = 90°

∴ ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

We are given,  

AM bisects ∠DAB i.e.,

∠DAM = ∠MAB = ½ ∠DAB = ½ * 90° = 45° …… (i)

And,  

DM bisects ∠CDA i.e.,

∠CDM = ∠MDA = ½ ∠CDA = ½ * 90° = 45° …… (ii)

Now, let’s consider ∆ AMD,  

∠AMD + ∠DAM + ∠MDA = 180° …. [∵ sum of all angles of triangle is 180°]

⇒ ∠AMD + 45° + 45° = 180° ……. [from (i) & (ii)]

⇒ ∠AMD = 180° - 45° - 45°  

⇒ ∠AMD = 90°

Hence proved

Answer 2

Answer:

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Step-by-step explanation:

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