Crossed Line Segments

Since $AE \times EB = CE \times ED = 240$ , by the converse of power of point theorem, quadrilateral ACBD is cyclic. Since ACBD is cyclic, by ptolemy theorem, $AC \times BD + BC \times AD = AB \times CD = (12 + 20) \times (10 + 24) = 1088$ , thus the last 3 digit is $088$ .

Observe that $\Delta ACE$ and $\Delta DBE$ are similar, and $\Delta AED$ and $\Delta CEB$ are similar. Let $x = AC$ and $y = BC$ . Then $BD = 1.2x$ and $AD=2y$ by ratio of similar triangles. Furthermore, $AC \times BD + BC \times AD = 1.2 x^2 + 2y^2.$

Let $\angle AEC = \theta$ . By the cosine law, we have $x^2 = 20^2 + 10^2 - 2 \times 20 \times 10 \cos \theta = 500 - 400 \cos \theta$ and $y^2 = 10^2 + 12^2 - 2 \times 10 \times 12 \cos (180 - \theta) = 244 + 240 \cos \theta .$ Hence, $AC \times BD + BC \times AD = 1.2 x^2 + 2y^2 = 1.2 (500 - 400 \cos \theta) + 2 (244 + 240 \cos \theta) = 1088.$

AE EB=20 12=240 CE ED=10 24=240. Thus, by the Power of a Point Theorem, AB and CD are chords of the same circle. Thus, ABCD is a cyclic quadrilateral.

Through Ptolemy's Theorem, AC BD+BC AD=AB CD in a cyclic quadrilateral. Since ABCD is cyclic, AC BD+BC AD=(20+12) (10+24)=1088. ITs last 3 digits is 088.

Note that AExEB=20x12=240=24x10=EDxEC, by Power of Point Theorem, ACBD is a cyclic quadrilateral. By Ptolemy's Theorem, ACxBD+BCxAD=ABxCD=34x32=1088 Therefore, the last three digits are 088

Note that ACBD is a cyclic quadrilateral since AE EB = CE ED. We can then apply Ptolemy's: AC BD+BC AD = AB CD. AB = AE+EB = 20+12 = 32, CD = CE+ED = 10+24 = 34. So, AB CD = 32 34 = 1088, and so is AC BD + BC*AD, so the last three digits are 088.

Since $20 \times 12=10 \times 24$ , we can use converse of Power of a Point to deduce that A, B, C, and D are concyclic. By Ptolemy's Theorem, we have that $AC \times BD+BC \times AD=AB \times CD$ . Thus, $AC \times BD+BC \times AD=(AE+BE)\times(CE+DE) = 32 \times 34 = 1088$ . Thus, the answer is 88 or 088.

Just giving suitable or comfortable values for the Vertices. you can choose any starting point on the X-Y coordinate Axes. I started with A =(0,0) and assumed AB on X axis(as no condition is given not to assume) and also AE = 20 implies point E=(20,0) so B becomes (32,0) as EB=12, similarly extending the same logic to CD(assuming AB perpendicular to CD and meeting at point E. we get D=(20,24), C=(20,10) Now get the values of AC,AD,BC,BD using distance formula and relate the terms and get the answer.

Yes, it is wrong. See here .

If the lines intersect perpendicularly, then we can find the sides AC,BD,BC and AD using Pythagoras theorem. Using Pythagoras theorem, AC = 10 $\sqrt{5}$ , AD = 4 $\sqrt{61}$ , BD = 12 $\sqrt{5}$ , and BC = 2 $\sqrt{61}$ . By substituting these values in the given expression, we get the answer as 1088. So, the last three digits were 088.

Let $AB$ and $CD$ be the chords of a circle with a radius $r$ . So, by Ptolemy's Theorem, we know

$AB \cdot CD = AC \cdot BD + BC \cdot AD$

Plug in the values, we will have $AC \cdot BD + BC \cdot AD = 32 \times 34 = 1088 \equiv \boxed{88} \pmod {100}$

$$ Since $\,AE\times EB=CE \times ED$ , then by using a cyclic quadrilateral's formula we can obtain that $$ $AC\times BD + BC \times AD = AB\times CD=1088.$ $$ Thus, the last 3 digits are $\;\boxed{088}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

AE times EB = CE times ED. This implies power of point and that line segs AB and CD are chords in a circle. These form diagonals of a cyclic quadrilateral and using Ptolemy's theorem. We obtain 1088 so the answer is 88