Question with math?

Question

There is a cyclic quadrilateral ABCD. Diagonals AC and BD intersect at point P, AB = AC = 6, angle BDC = 1/2 of angle BAC. If CP = 1, find BP times DP.

I'm not sure how to solve this.

Answer 1

By the intersecting chords theorem (Power of a Point):

AP×PC=BP×PD. Since AC=6 and CP=1 then:

AP=6−1=5 Therefore,

BP×PD=AP×PC=5×1=5 Answer:

5

Answer 2

Let's start with what we know about the problem:

1. The shape is a cyclic quadrilateral. This means that all 4 vertices fall on a single circle. It also means the following: ∠ABC + ∠ADC = 180 and ∠BCD + ∠BAD = 180 (opposite angles are supplementary)
2. AB = AC = 6. This means further that ∠ABC = ∠ACB, because we can now say that triangle ABC is isosceles.
3. ∠BAC = 2 * ∠BDC
4. CP = 1

Let's say that ∠BDC = x. This makes ∠BAC = 2x. Let's also say that ∠ABC = ∠ACB = y. We can now say that the inscribed angles of triangle ABC sum to 180, or that ∠ABC + ∠ACB + ∠BAC = 180, meaning that y + y + 2x = 180. We can then solve for y to say that y = 90 - x.

Now, ∠ABC + ∠ADC = 180; ∠ABC = y = 90 - x; and ∠ADC = ∠ADB + ∠BDC = ∠ADB + x.

We can therefore combine those equations to say that ∠ADB + x + (90 - x) = 180, which tells us that ∠ADB = 90 (it is a right angle).

Here's the trick to the problem: if the angle ∠ADB is a right angle, then the segment AB bisects the circle that contains all four vertices (it is a diameter of the circle). This means that ∠ACB is also a right angle.

But, since triangle ABC is isosceles and thus ∠ABC = ∠ACB = 90, we would therefore conclude that the angle ∠BAC is zero, that points B and C are co-located, and that therefore point P would also be co-located with point B and C, and CP thus cannot equal 1.

YOU GOT AN IMPOSSIBLE PROBLEM: THERE IS NO ANSWER

Answer 3

Since ABCD is a cyclic quadrilateral, we can use several important facts about angles and intersecting chords.

We are told:

• ABCD is cyclic

• AB = AC = 6

• ∠BDC = (1/2) ∠BAC

• AC and BD intersect at P

• CP = 1

We want to find BP × DP.

First, use the angle information. In a cyclic quadrilateral, inscribed angles that subtend the same arc are equal.

Here, ∠BAC subtends arc BC.

∠BDC also subtends arc BC.

So normally ∠BDC = ∠BAC.

But the problem states ∠BDC = (1/2) ∠BAC, which means point D is located so that triangle BDC is formed by “bisecting” the arc position.

This special configuration implies that point D is chosen so that BD is the symmedian of triangle ABC.

A key property of symmedians is that when BD intersects AC at P:

(AP / PC) = (AB² / CB²).

We know AB = AC = 6, so triangle ABC is isosceles with AB = AC.

Let BC = x.

Then:

AP / PC = AB² / BC² = 36 / x².

We are given CP = 1, so:

AP = 36 / x².

Next, apply the intersecting chords theorem on diagonals AC and BD:

(AP)(PC) = (BP)(PD).

We already know:

PC = 1

AP = 36 / x²

So:

BP × DP = (AP)(PC) = (36 / x²)(1) = 36 / x².

To finish, we use the condition ∠BDC = (1/2) ∠BAC, which forces D to lie on the symmedian, and in this specific symmetric case (AB = AC), the construction leads to BC = 6 as well (the classical equal-leg symmetric configuration of the problem).

Thus x = BC = 6, giving:

BP × DP = 36 / 36 = 1.

Final Answer: BP × DP = 1.

Answer 4

There is a theorem called the intersecting chords theorem. Remember that a chord is any line segment where each end of the segment is on the circumference of the circle. If two chords intersect as they cross the circle, multiplying the two segments of one chord will equal the product of the two segments of the other chord.

In this problem, AP × PC = DP × PB as those are the intersecting chords broken into segments where they cross. All of AC = 6, and its smaller segment PC = 1. That means the larger segment of AP = 5.

Put it all together and AP × PC = 5 × 1 = 5. That’s also the answer to question of DP × PB.

The strange part of this question is the inclusion of the angle information. ∠BAC and ∠BDC should be congruent because they are inscribed angles that intersect the same arc. So they should both be half the measurement of the BC arc. One can’t be half the size of the other.