Advertisements
Advertisements
Question
ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Advertisements
Solution
Given: ABCDE is a regular pentagon.
The bisector ∠A of the pentagon meets the side CD at point M.
To prove : ∠AMC = 90°
Proof: We know that the measure of each interior angle of a regular pentagon is 108°.
∠BAM = 
Since, we know that the sum of a quadrilateral is 360°
In quadrilateral ABCM, we have
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
54° + 108° + 108° + ∠AMC = 360°
∠AMC = 360° – 270°
∠AMC = 90°
APPEARS IN
RELATED QUESTIONS
Complete of the following, so as to make a true statement:
A quadrilateral has ....... sides.
Complete of the following, so as to make a true statement:
A quadrilateral has ..... vertices, no three of which are .....
In Fig. 16.19, ABCD is a quadrilateral.
Name a pair of adjacent sides.
In Fig. 16.19, ABCD is a quadrilateral.
How many pairs of adjacent sides are there?
In Fig. 16.21, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
The angles of a hexagon are (2x + 5)°, (3x - 5)°, (x + 40)°, (2x + 20)°, (2x + 25)° and (2x + 35)°. Find the value of x.
The number of right angles in a straight angle is ______ and that in a complete angle is ______.
What conclusion can be drawn from part of given figure, if BD bisects ∠ABC?
Draw a rough sketch of a quadrilateral KLMN. State two pairs of adjacent sides.
