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In the triangle ABC, AB=AC.If the perpendicular distances to AB and AC from any point O on BC are OP and OQ respectively and the perpendicular distance to AC from B is BD. Prove that OP OQ=BD.​?
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In the triangle ABC, AB=AC.If the perpendicular distances to AB and AC...
Proof:

Given: In triangle ABC, AB=AC. Perpendicular distances to AB and AC from any point O on BC are OP and OQ respectively, and the perpendicular distance to AC from B is BD.

To prove: OP * OQ = BD

Proof:

Step 1: Draw a rough diagram

Let's start by drawing a rough diagram of triangle ABC:

```
B
/ \
BD \
/ \
A-------C
```

Step 2: Mark the given information

- AB = AC (Given)
- Perpendicular distance from O to AB is OP
- Perpendicular distance from O to AC is OQ
- Perpendicular distance from B to AC is BD

Step 3: Draw perpendiculars from O and B

Now, let's draw perpendiculars from point O and point B to AB and AC respectively:

```
B
/ \
BD \
/ \
A-------C
| \
| \
O \
| \
| \
P Q
```

Step 4: State the theorem to be proven

We need to prove that OP * OQ = BD.

Step 5: Prove OP * OQ = BD

Since AB = AC (Given), the perpendicular distances from O to AB and AC are equal (OP = OQ) because O lies on the perpendicular bisector of BC.

Therefore, OP * OQ = OQ * OQ (Substituting OP with OQ)

= OQ^2

Also, since B is perpendicular to AC, BD is the perpendicular distance from B to AC.

Therefore, OP * OQ = BD^2 (Substituting OQ^2 with BD^2)

Hence, OP * OQ = BD.

Step 6: Conclusion

We have proved that OP * OQ = BD using the given information and properties of a triangle.

Therefore, the statement is true.
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In the triangle ABC, AB=AC.If the perpendicular distances to AB and AC from any point O on BC are OP and OQ respectively and the perpendicular distance to AC from B is BD. Prove that OP OQ=BD.​?
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