Question

In the given figure, PQR is an equilateral triangle, AB ║ PR and PR is produced to D such that RD = QA. Prove that BD bisects AR.

Answer 1

Answer:

Mark as brainliest.

Hope it is helpful

Step-by-step explanation:

It is given that ABC is an equilateral triangle; PQ∥AC and AC is produced to R such that CR=BP

Consider QR intersecting the line PC at point M

We know that △ABC is an equilateral triangle

So we get ∠A=∠ACB=60

∘

From the figure we know that PQ∥AC and ∠BPQ and ∠ACB are corresponding angles

So we get

∠BPQ=∠ACB=60

∘

Based on the △BPQ we know that

∠B=∠ACB=60

∘

It can be written as

∠BQP=60

∘

According to the figure we know that △BPQ is an equilateral triangle

So we get

PQ=BP=BQ

It is given that CR=BP so we get

PQ=CR.(1)

In the △PMQ and △CMR we know that PQ∥AC and QR is the transversal

We know that ∠PQM and ∠CRM are alternate angles and ∠PMQ and ∠CMR are vertically opposite angles

∠PQM=∠CRM

∠PMQ=∠CMR

By considering equation (1) and AAS congruence criterion

△PMQ≅△CMR

We know that the corresponding parts of congruent triangles are equal

PM=MC

Therefore, it is proved that QR bisects PC.

Answer 2

Given :

• ∆PQR is an equilateral triangle
• AB ║ PR
• RD = QA.

To Find :

• BD bisects AR

Solution :

• Since we are given that , ∆PQR is an equilateral triangle. Therefore all sides of ∆PQR will be equal to 60°. mathematically ;

→∠P + ∠Q + ∠R = 60°

• Now, we are also provided that AB ║ PR Therefore :

→∠P = ∠R [Corresponding angles]

∴ ∠P = 60°

• So, by observation we found that ∠P and ∠Q is equal to 69° therefore, the third side of the ∆ will also be 60°.

→∠Q + ∠A + ∠B = 60°

∴ ∆QAB is an equilateral traingle

→QB = AB

→QA = RD

→∠R = ∠M [Alternate angels]

∴ ∆ABM ≅ ∆RDM [AAS Property]

∴ AM = RM

∴BD bisects AR.

Hence Proved !