## 10th Grade Geometry Solution5

In this page 10th grade geometry solution5 we are going to see solutions of some practice questions.

(11) In a triangle MNO, MP is the external bisector of angle M meeting NO produced at P. IF MN = 10 cm, MO = 6 cm, NO - 12 cm, then find OP.

Solution: MP is the external bisector of angle M

by using angle bisector theorem in the triangle MNO we get,

(NP/OP) = (MN/MO)

NP = NO + OP

= 12 + OP

(12 + OP)/OP = 10/6

6 (12 + OP) = 10 OP

72 + 6 OP = 10 OP

72 = 10 OP - 6 OP

4 OP = 72

OP = 72/4

= 18 cm

(12) In a quadrilateral ABCD, the bisectors of ∠B and ∠D intersect on AC at E. Prove that (AB/BC) = (AD/DC).

Solution: Here DE is the internal angle bisector of angle D.

by using internal bisector theorem, we get

(AE/EC) = (AD/DC)  ----- (1)

Here BE is the internal angle bisector of angle B.

(AE/EC) = (AB/BC)  ----- (2)

from (1) and (2) we get

Hence proved

(13) The internal bisector of ∠A of triangle ABC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that (BD/BE) = (CD/CE).

Solution: 10th grade geometry solution5 10th grade geometry solution5

In triangle ABC, AD is the internal bisector of angle A.

by using angular bisector theorem in triangle ABC

(BD/DC) = (AB/AC) ----- (1)

In triangle ABC, AE is the internal bisector of angle A

(BE/CE) = (AB/AC) ----- (2)

from (1) and (2) we get,

(BD/DC) = (BE/CE)

(BD/BE) = (DC/CE)

Hence proved

(14) ABCD is a quadrilateral with AB = AD. If AE and AF are internal bisectors of ∠BAC and ∠DAC respectively,then prove that the sides EF and BD are parallel.

Solution: In triangle ABC,AE is the internal bisector of ∠BAC

by using bisector theorem, we get

(AB/AC) = (BE/EC)  ----- (1)

In triangle ADC,AF is the internal bisector of ∠DAC

by using bisector theorem, we get

Since lengths of AD and AB are equal, we are going to replace AB instead of AD

(AB/AC) = (DF/FC)  ----- (2)

from (1) and (2) we get

(DF/FC) = (BE/EC)

So we can decide EF and BD are parallel by using converse of "Thales theorem".  