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Fittoniya [83]
3 years ago
15

A line segment BK is an angle bisector of ΔABC. A line KM intersects side BC such, that BM = MK. Prove: KM ∥ AB.

Mathematics
2 answers:
weqwewe [10]3 years ago
8 0

Answer:

∠BKM= ∠ABK

Therefore AB ║KM (∵ ∠BKM= ∠ABK  and lies between AB and KM and BK is the transversal line)

m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)

Step-by-step explanation:

Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK

TO prove: KM ║AB

Now, As given in figure 1,

In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)

Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)

Now ∵ ∠MBK = ∠BKM

and  ∠ABK = ∠KBM

∴ ∠BKM= ∠ABK

Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)

Hence proved.  

Serga [27]3 years ago
7 0

Answer:

∠BKM= ∠ABK

Therefore AB ║KM (∵ ∠BKM= ∠ABK  and lies between AB and KM and BK is the transversal line)

m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)

Step-by-step explanation:

Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK

TO prove: KM ║AB

Now, As given in figure 1,

In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)

Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)

Now ∵ ∠MBK = ∠BKM

and  ∠ABK = ∠KBM

∴ ∠BKM= ∠ABK

Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)

Hence proved.  

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