Parallelogram ABCD: Finding AC from PB & PD

In summary, a parallelogram is a four-sided shape with opposite sides that are parallel and equal in length. To find the length of side AC in parallelogram ABCD, you can use the Pythagorean Theorem. The properties of a parallelogram include four sides, four angles, two pairs of parallel sides, and diagonals that bisect each other and make equal angles with the sides. It is possible to find the length of side AC if you know the lengths of PB and PD using the Pythagorean Theorem. Lastly, a parallelogram differs from a rectangle in that a rectangle has four right angles and all equal sides, while a parallelogram can have any angle measure and its opposite sides are equal
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In parallelogram $ABCD$, $\angle B$ and $\angle D$ are acute while $\angle A$ and $\angle C$ are obtuse. The perpendicular from $C$ to $AB$ and the perpendicular from $A$ to $BC$ intersect at $P$ inside the parallelogram. If $PB=700$ and $PD=821$, find $AC$.
 
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[TIKZ]
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:C] (C) at (12, 0);
\coordinate[label=above:A] (A) at (3,10);
\coordinate (F) at (3,0);
\coordinate (E) at (0.991,3.303);
\coordinate[label=above: P] (P) at (3,2.7);
\coordinate[label=above: D] (D) at (15,10);
\draw (A) -- (B)-- (C)-- (D) -- (A);
\draw (C) -- (E);
\draw (A) -- (F);
\draw (P) -- (D);
\draw (F) rectangle +(-0.2, 0.2);
\draw[thick,dashed] (9,6.35) circle (7.023cm);
\begin{scope}[shift={(1,0)}]
\node[draw,rectangle,rotate=71] at (0.16,3.395){};
\end{scope}
[/TIKZ]

First note that $P$ is the orthocenter of $\triangle ABC$. Furthermore, note that from the perpendicularity $DA\perp AP$ and $DC\perp CP$, so quadrilateral $DAPC$ is cyclic. Furthermore, $DP$ is a diameter of circle $(DAPC)$. This is the circumcircle of $\triangle DAC$, which is congruent to $BCA$. As a result, if $R$ is the circumradius of $\triangle ABC$, then $PD=2R$.

Now I claim that $PB=2R\cos B$. To prove this, reflect $P$ across $AB$ to point $P'$. It is well-known that $P'$ lies on the circumcircle of $\triangle ABC$, so in particular the circumradii of $\triangle APB$ and $\triangle ACB$ are equal. But then by Law of Sines \[\dfrac{BP}{\sin\angle BAP}=\dfrac{BP}{\cos B}=2R\quad\implies\quad BP = 2R\cos B\]as desired. (An alternate way to see this is through the diagram itself: from right triangle trigonometry on triangles $DAP$ and $DCP$ it is not hard to see that $PA=2R\cos A$ and $PC=2R\cos C$, which by symmetry suggests $PB=2R\cos B$.)

Finally, note that by Law of Sines again we have $AC=2R\sin B$, so \[AC^2+BP^2=(2R\sin B)^2 + (2R\cos B)^2 = (2R)^2(\sin^2 B+\cos^2 B) = PD^2.\]Hence \[AC^2=PD^2-PB^2=821^2-700^2=(821-700)(821+700)=11^2\cdot 39^2\]and so $AC=11\cdot 39=\boxed{429}$.
 

FAQ: Parallelogram ABCD: Finding AC from PB & PD

What is a parallelogram?

A parallelogram is a four-sided shape with opposite sides that are parallel and equal in length. It also has opposite angles that are equal in measure.

How do you find the length of AC in parallelogram ABCD?

To find the length of AC, you can use the formula AC = PB + PD, where PB and PD are the lengths of the diagonals of the parallelogram.

Can you find the length of AC if you only know the lengths of PB and PD?

Yes, you can find the length of AC using the formula AC = PB + PD, as long as you know the lengths of both diagonals of the parallelogram.

Is there another way to find the length of AC in parallelogram ABCD?

Yes, you can also use the Pythagorean theorem to find the length of AC. This involves using the lengths of the sides of the parallelogram to find the length of the diagonal, which is equal to AC.

Why is finding the length of AC important in parallelogram ABCD?

Finding the length of AC can help us determine the area and perimeter of the parallelogram, as well as other properties such as the angles and side lengths. It is an important step in understanding and analyzing the shape.

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