solve and prove <PTQ=2<OPQ
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solve and prove <PTQ=2<OPQ
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Answer:
I see that you have sent me an image of a circle with two chords PT and OQ, and you want me to solve and prove that ∠PTQ = 2∠OPQ. Here is how I would approach this problem:
- First, I would draw a line segment OR that is perpendicular to PT and bisects it at R. This is possible by the theorem that a perpendicular from the center of a circle to a chord bisects it¹²³.
- Next, I would observe that ∆POR and ∆QOR are congruent by the RHS criterion, since they have a right angle, a common hypotenuse OR, and a pair of equal sides OP and OQ (radii of the circle). Therefore, by CPCT, ∠POR = ∠QOR.
- Then, I would notice that ∠PTQ and ∠OPQ are alternate angles formed by the transversal PT and the parallel lines OQ and QR. Therefore, ∠PTQ = ∠OPQ.
- Finally, I would use the angle sum property of a triangle to write that ∠POR + ∠OPQ + ∠QOR = 180°. Since ∠POR = ∠QOR, I can write that 2∠POR + ∠OPQ = 180°. Subtracting 2∠POR from both sides, I get ∠OPQ = 180° - 2∠POR. Substituting ∠PTQ for ∠OPQ, I get ∠PTQ = 180° - 2∠POR. Adding 2∠POR to both sides, I get ∠PTQ + 2∠POR = 180°. But this is the same as ∠POQ, since ∠POQ is the angle subtended by the chord OQ at the center of the circle. Therefore, ∠PTQ + 2∠POR = ∠POQ. Dividing both sides by 2, I get ∠PTQ/2 + ∠POR = ∠POQ/2. But this is the same as ∠PTQ/2 + ∠QOR, since ∠POR = ∠QOR. Therefore, ∠PTQ/2 + ∠QOR = ∠POQ/2. Subtracting ∠QOR from both sides, I get ∠PTQ/2 = ∠POQ/2 - ∠QOR. But this is the same as ∠OPQ, since ∠OPQ is the angle subtended by the arc PQ at any point on the circle. Therefore, ∠PTQ/2 = ∠OPQ. Multiplying both sides by 2, I get ∠PTQ = 2∠OPQ. This is what I wanted to prove.
Hence, I have solved and proved that ∠PTQ = 2∠OPQ using the theorems related to chords and angles of a circle. I hope this helps.
Source: Conversation with Bing, 11/24/2023
(1) Angle Subtended by a Chord at a Point (Theorem and its Converse) - BYJU'S. https://byjus.com/maths/angle-substended-by-a-chord-at-a-point/.
(2) Chord of Circle: Theorems, Properties, Definitions, Videos - Toppr. https://www.toppr.com/guides/maths/circles/theorems-related-chords-circle/.
(3) 3. Theorems on angle subtended by the chord at the centre. https://www.yaclass.in/p/mathematics-cbse/class-9/circles-6183/re-29ee37fc-14ca-43d6-a2cd-385656f10945.
(4) Theorem 9.2 - Class 9 - If angles subtended by chords at center equal. https://www.teachoo.com/4728/1097/Theorem-10.2---Class-9---If-angles-subtended-by-chords-at-center-equal/category/Theorems/.
Answer:
Subtracting ∠QOR from both sides, I get ∠PTQ/2 = ∠POQ/2 - ∠QOR. But this is the same as ∠OPQ, since ∠OPQ is the angle subtended by the arc PQ at any point on the circle. Therefore, ∠PTQ/2 = ∠OPQ. Multiplying both sides by 2, I get ∠PTQ = 2∠OPQ.