Here are my submissions :
1) Let \( \Delta ABC \) have cevians \( \overline{AD} \) and \( \overline{CE} \), which meet at a point \( F \) inside the triangle. Prove that \( [\Delta ABC] \cdot [\Delta DEF] = [\Delta BDE] \cdot [\Delta AFC] \), where \( [ A ] \) denotes area of figure \( A \).
2) If a cevian of an equilateral triangle is extended to meet the circumcircle at , prove that .
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Question 2: We know by Ptomely's, that AP=PB+PC.
Then we just need to prove that PB1+PC1=PB⋅PCAP=PQ1⟹APBP=CPPQ
Observe that ∠ABP=∠ACB=60∘ and ∠APC=∠ABC=60∘ because they substend the same arc. Similarly, ∠CBP=∠CAP substend the same arc. This implies that △BPQ∼APC and the desired ratio is given with these proportional lengths.
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@Alan Yan Do check problem no. 5 in RMO board-2 by Nihar Mahajan ! It's more interesting than these :). Link
Question 1: Extend segment BF to hit AC at T. Construct DE and call its intersection point with BE , M. Notice that the desired relation is equivalent to [BDE][DEF]=[ABC][AFC]⟹MBFM=TBTF. We will use mass points. Put weights of p,q,r on A,B,C respectively. This implies that T has a mass of p+r with implies that BFTF=p+rq⟹TBTF=p+q+rq
Notice that F has mass p+q+r. This implies that MBMF=p+q+rq and we are done.
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Nice use of Barycentric Coordinates !
@Xuming Liang @Calvin Lin @Sualeh Asif Here are some simple problems ! Presently not very good at Geometry, but planning to improve.