In the parallelogram ABCD, a line through A meets BD at P, CD at Q and BC extended at R. Prove the PQ/PR = (PD/PB)2.
∠PAB = ∠PQD by alternate angles.
∠PBA = ∠PDQ by alternate angles.
Therefore, trianglesPAB and PQD are similar because they have the same angles.
So PA/PQ= PB/PD.
∠PAD = ∠PRB by alternate angles.
∠PDA = ∠PBR by alternate angles.
Therefore, triangles PAD and PRB are similar because they have the same angles.
So PA/PD= PR/PB.
These equations give us PA = PR × PD / PB = PB × PQ / PD.
This rearranges to give PQ/PR = (PD/PB)2
This course guides you through the fundamentals of Python programming using an interactive Python library known as Turtle.
This course encompasses a range of Geometry topics such as coordinate and spatial geometry, introductory trigonometry, angles, parallel lines, congruent and similar triangles, polygons, circles, the Pythagorean Theorem, and more. Emphasis will be placed on reinforcing Algebra skills and enhancing critical thinking through problem-solving in both mathematical and real-world contexts.
Ask about our courses and offerings, and we will help you choose what works best for you.